A spring of spring constant $$5 \times {10^3}N/m$$   is stretched initially by $$5\,cm$$  from the unstretched position. Then the work required to stretch it further by another $$5\,cm$$  is

Solution :
$$\eqalign{ & {W_1} = \frac{1}{2} \times 5 \times {10^3}{\left( {0.05} \right)^2} \cr & \Rightarrow {W_2} = \frac{1}{2} \times 5 \times {10^3}{\left( {0.10} \right)^2} \cr & \therefore \Delta W = \frac{1}{2} \times 5 \times {10^3} \times 0.15 \times 0.05 = 18.75\,J. \cr} $$